Question

The mean rent of a 3-bedroom apartment in Orlando is $1300. You randomly select 10 apartments around town. The rents are normally distributed with a standard deviation of $350. What is the probability that the mean rent is more than $1100?

Answer 1

96.49% probability that the mean rent is more than $1100

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

`\mu = 1300, \sigma = 350, n = 10, s = (350)/(√(10)) = 110.68`

What is the probability that the mean rent is more than $1100?

This is 1 subtracted by the pvalue of Z when X = 1100. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (1100 - 1300)/(110.68)`

`Z = -1.81`

`Z = -1.81` has a pvalue of 0.0351

1 - 0.0351 = 0.9649

96.49% probability that the mean rent is more than $1100